The number of four-digit numbers that can be formed by using the digits `1,2,3,4,5,6,7,8` and `9` such that the least digit used is `4`, when repetiti

Correct option is: C. 18

Correct Answer - B The prime digits are 2, 3, 5, 7. If we fix 2 at first place, then other 2n - 1 places are filled by all four digits....

Correct Answer - D Three-digit numbers are 100, 101, …, 999. Total number of such numbers is 900. The three-digit numbers (which have all same digits) are 111, 222, 333, …,...

We have `17^(256) = (17^(2))^(128) = (289)^(128) = (290-1)^(128)` `:. 17^(256) = .^(128)C_(0)(290)^(128)-.^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `- .^(128)C_(125)(290)^(3)+.^(128)C_(126)(290)^(2)-.^(128)C_(127)(290)+1` `=[.^(128)C_(0)(290)^(128) - .^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `-.^(128)C_(125)(290)^(3)]+.^(128)C_(126)(290)^(2) -.^(128)C_(127)(290)+1` `=1000m+((128)(127))/(2)(290)^(2)-128xx290+1` `= 1000 m + (128)(127)(290)(145)-128xx290-1` `= 1000m+(128)(290)(127xx145-1)+1` `=1000m+(128)(290)(18414)+1` `=1000m+683527680+1`...

Correct Answer - `1-(1-p)^(8)` Probability for an incorrec digit is p. hence, probability for 8 correct digit is `1-p)^(8).` Hence, required probability is `1-(1-p)^(8).`

Correct Answer - C `(c )` The sum of digits must be divisible by `9` The pair of missing numbers must be `(1,2)` or `(0,3)` Case I : `(1,2)` `implies` when...

Correct Answer - B `(b)` `12=2^(2)*3` Then the four digits can be `1,1,6,2 to` no. of numbers `=(4!)/(2!)=12` `1,1,3,4 to` no. of numbers `=(4!)/(2!)=12` `2,2,3,1 to` no. of numbers `=(4!)/(2!)=12` Hence,...

Correct Answer - C `(c )` Case I : All six digits alike i.e. `111111`, `222222`……..etc. `=5` ways Case II : `2` alike `+2` other alike. Select any three in `"^(5)C_(3)`...

Correct Answer - B `(b)` Required number of numbers `="(n)C_(0)(2)^(n)+"(n)C_(2)(2)^(n-2)+…` We know that `(1+x)^(n)+(x-1)^(n)=2sum_(k=0)^(n)"(n)C_(k)x^(k)` Hence, put `x=2`, get the result `2["^(n)C_(0)(2)^(n)+^(n)C_(2)(2)^(n-2)+^(n)C_(2)(2)^(n-4)+^(n)C_(6)(2)^(n-6)+...]`

Correct Answer - B `(b)` Total number of ways such that at least `3` digits will not occur in its position. `=^(5)C_(3){3!-^(3)C_(1)2!+^(3)C_(2)1!-^(3)C_(2)0!}+^(5)C_(4)[4!-^(4)C_(1)(3!)+^(4)C_(2)(2!)-^(4)C_(3)(1!)+^(4)C_(4)(0!)}+^(5)C_(5){5!-^(5)C_(1)4!+^(5)C_(2)3!-^(5)C_(3)2!+^(5)C_(4)1!-^(5)C_(5)(0!)}` `=10(2)+5(9)+(44)` `=20+45+44=109`